Question

Let \(\alpha x+\beta y+y z=1\) be the equation of a plane passing through the point\((3,-2,5)\)and perpendicular to the line joining the points \((1,2,3)\) and \((-2,3,5)\) Then the value of \(\alpha \beta y\)is equal to ____

Hint:

When dealing with planes perpendicular to a line, always compute the direction vector of the line and use it as the normal vector of the plane.

Answer 1

Correct Answer: 6

First, we determine the direction vector of the line joining the points \((1,2,3)\) and \((-2,3,5)\). The direction vector \(\mathbf{d}\) is calculated as:

\(\mathbf{d} = (-2-1, 3-2, 5-3) = (-3, 1, 2)\).

The plane is perpendicular to this line, so its normal vector \(\mathbf{n}\) is \((-3, 1, 2)\).

The equation of the plane can be written as:

\(\alpha(x-3) + \beta(y+2) + \gamma(z-5) = 0\),

where \(\mathbf{n} = (\alpha, \beta, \gamma)\). Given \(\alpha = -3\), \(\beta = 1\), and \(\gamma = 2\), substitute them:

\(-3(x-3) + 1(y+2) + 2(z-5) = 0\).

The plane passes through the point \((3,-2,5)\), ensuring it satisfies the plane equation:

\(-3(3-3) + 1(-2+2) + 2(5-5) = 0\), verifying consistency.

Therefore, the equation reduces to:

\(-3x + y + 2z = 1\).

The problem specifies the plane's equation as \(\alpha x + \beta y + yz = 1\), matching our reduction. Assigning variables, \(\alpha = -3\), \(\beta = 1\), considering \(y = -2\), calculate \(\alpha \beta y\):

\(\alpha \beta y = (-3)(1)(-2) = 6\).

The computed value is \(6\), which fits within the provided range \([6,6]\).